Cosine Problem for Trig

Question

if y= -cos x, what x-value corresponds to a y-value of 1 between 0 and 2 pi

Answer 1

The Cosine Function takes on a value of 1 at X= 0 and x= 2π. it takes on values of 0 at

x= π/2 and x=3π/2. it takes on a value of minus 1 at x=π or Cos(π) ==1. so, -cos(π)= 1

The answer to your question is y=1 when x= π

Hope this helps

Answer 2

For this problem, we're given a relationship between y and x. We are told that for a given value of x, the negative of the cosine value of that x is equal to y. This is the verbal form of the equation "y = - cos(x)".

The question then asks us what the value of x will be if our y-value is 1, and they want our answer for x-value to be in between 0 and 2 pi radians.

(If this radian business doesn't make any sense, I recommend searching for a basic rundown of radians in trigonometry. Briefly, radians are a unit that we apply instead of "degrees" when measuring angles in geometry because they incorporate the detail of pi in describing the relationship between sine, cosine, and tangent ratios.)

So how do we get the equation "y = - cos(x)" to show us x by itself? Well, first we move our negative sign over to the y, and then we take the inverse cosine of each side. You'll see this written as "csc", "cos^-1", or cosecant in the long form. Most often we write it as cos^-1 because that most clearly demonstrates the inverse nature of this function. So we end up with "x = cos^-1 (-y)", and then we have to plug in our y-value of 1 as stated in the problem.

This gives us a value of -1.1 radians (plus a lot more digits!), which is outside the range of 0 to 2 pi. To get this value into the range requested by the problem, we simply have to add 2 pi. This is effectively the same as walking all the way around a circle to the same spot. Our endpoint describes the same relationship angle back to the center/zero radian point, but the radian value is in our requested range. Thus, our end value is -1.1 + 2 pi radians, which comes out to 5.0948 radians (or 1.6217 pi radians).

Hope this helps -- ask any follow up questions if you have them!